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Ch.13 - Properties of Solutions

Chapter 13, Problem 106a

Fluorocarbons (compounds that contain both carbon and fluorine) were, until recently, used as refrigerants. The compounds listed in the following table are all gases at 25 °C, and their solubilities in water at 25 °C and 1 atm fluorocarbon pressure are given as mass percentages. (a) For each fluorocarbon, calculate the molality of a saturated solution.

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Hi everyone here we have a question telling us the following are some sulfate salts with their cell viability. Mass percent and water at 25 degrees Celsius, silver sulfate equals 0.830 blood sulfate equals 0.44 Or three. Barium sulfate equals 0.245. And strontium sulfate equals 0.0135. If a saturated solution was created individually for each of the sulfate salts given above, what will be their morality. So solly ability and water equals grams of solute Over 100 g of water. So let's start out here with silver sulfate. We have zero .83g times one mole divided by It's Mueller mass 311.799 g divided by g of water times one kg, Divided by 1000 g equals zero point zero 2662. And if we want to put that in scientific notation we're gonna move the decimal point to the right 12 places. So that is 2. Times 10 to the -2. And that is our first answer. Now we're going to do the same for lead. Sulfate, not equals 0.0044, three grams times one mole, Divided by 303 . g divided by 100 g of water times one kg Divided by g and that equals 1. Times 10 to the -4. And that is our second answer. Next we have barium sulfate which equals 0. times one mole Divided by its molar mass. So 233.38 g divided by 100 g of water, times one kg Divided by g. And that equals 0. 49 79. And if we want to put that in scientific notation, we're gonna move our decimal point to the right 12 places. So that would be 1. times 10 to the negative two. And that is our third answer. And then lastly we have strontium sulfate which equals 0.0135 g times one mole Divided by its molar mass. 183 .68 g Divided by 100 g of water times one kg Divided by 1000 g. And that equals 7. Times 10 to the -4. And that is our last answer. Thank you for watching. Bye.
Related Practice
Textbook Question

Two beakers are placed in a sealed box at 25 °C. One beaker contains 30.0 mL of a 0.050 M aqueous solution of a nonvolatile nonelectrolyte. The other beaker contains 30.0 mL of a 0.035 M aqueous solution of NaCl. The water vapor from the two solutions reaches equilibrium. (a) In which beaker does the solution level rise, and in which one does it fall?

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Textbook Question

Two beakers are placed in a sealed box at 25 °C. One beaker contains 30.0 mL of a 0.050 M aqueous solution of a nonvolatile nonelectrolyte. The other beaker contains 30.0 mL of a 0.035 M aqueous solution of NaCl. The water vapor from the two solutions reaches equilibrium. (b) What are the volumes in the two beakers when equilibrium is attained, assuming ideal behavior?

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Textbook Question

Carbon disulfide (CS2) boils at 46.30 °C and has a density of 1.261 g/mL. (a) When 0.250 mol of a nondissociating solute is dissolved in 400.0 mL of CS2, the solution boils at 47.46 °C. What is the molal boiling-point-elevation constant for CS2?

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Textbook Question

Fluorocarbons (compounds that contain both carbon and fluorine) were, until recently, used as refrigerants. The compounds listed in the following table are all gases at 25 °C, and their solubilities in water at 25 °C and 1 atm fluorocarbon pressure are given as mass percentages. (c) Infants born with severe respiratory problems are sometimes given liquid ventilation: They breathe a liquid that can dissolve more oxygen than air can hold. One of these liquids is a fluorinated compound, CF3(CF2)7Br. The solubility of oxygen in this liquid is 66 mL O2 per 100 mL liquid. In contrast, air is 21% oxygen by volume. Calculate the moles of O2 present in an infant’s lungs (volume: 15 mL) if the infant takes a full breath of air compared to taking a full “breath” of a saturated solution of O2 in the fluorinated liquid. Assume a pressure of 1 atm in the lungs.

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Textbook Question

At ordinary body temperature (37 °C), the solubility of N2 in water at ordinary atmospheric pressure (1.0 atm) is 0.015 g/L. Air is approximately 78 mol % N2. (b) At a depth of 100 ft in water, the external pressure is 4.0 atm. What is the solubility of N2 from air in blood at this pressure?

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Textbook Question
A series of anions is shown below:

The anion on the far right is called 'BARF' by chemists, as its common abbreviation sounds similar to this word. (d) Tetrabutylammonium, (CH3CH2CH2CH2)4N + is a bulky cation. Which anion, when paired with the tetrabutylammonium cation, would lead to a salt that will be most soluble in nonpolar solvents?
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